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Brandon invested $9,200 in an account paying an interest rate of 3 1/4% compounded quarterly. Lamonte invested $9,200 in an account paying an interest rate of 2 7/8 % compounded monthly. After 19 years, how much more money would Brandon have in his account than Lamonte, to the nearest dollar?

2 Answers

1 vote

To solve this problem, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

where A is the amount of money at the end of the investment period, P is the principal (initial investment), r is the interest rate (expressed as a decimal), n is the number of times the interest is compounded per year, and t is the number of years of the investment period.

For Brandon's account:

P = $9,200

r = 3.25% = 0.0325

n = 4 (compounded quarterly)

t = 19 years

A = 9200(1 + 0.0325/4)^(4*19) = $17,631.51

For Lamonte's account:

P = $9,200

r = 2.875% = 0.02875

n = 12 (compounded monthly)

t = 19 years

A = 9200(1 + 0.02875/12)^(12*19) = $16,031.63

The difference in the amount of money between Brandon's and Lamonte's accounts after 19 years is:

$17,631.51 - $16,031.63 = $1,599.88

Therefore, Brandon would have approximately $1,599 more in his account than Lamonte after 19 years, to the nearest dollar.

User Charles Anthony
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1 vote

Answer:1141

Explanation:

User Zsuzsa
by
7.8k points